The trough normal incidence efficiency from combining collector optical efficiency and receiver heat loss results is weakly dependent on the radiation values in Watt per square meters (above 350 degrees Celsius).

For example, at 100 degrees Celsius, the efficiency is about 75%. At 200 degrees Celsius, the efficiency is about 73%.

Here is a nice report for you from the following URL:

http://www.nrel.gov/docs/fy11osti/49304.pdf

You download it free from www.NREL.GOV

Hope this helps answer your question.

Professor Yehia Khalil

Yale University

USA

The optical analysis is basic because the performance of teh collector is fully connected with this. First of all, it is important to understand that the solar collector have to follow the sun via a tracking system. The tarcking system aims to move the parabola in order to minimize the transevrsal angle of the solar irradiation. In order to understand this concept, you can read the nexta publication:

Y. Wang, Y. Zhu, H. Chen, X. Zhang, L. Yang, C. Liao, Performance analysis of a novel sun-tracking CPC heat pipe evacuated tubular collector, Applied Thermal Engineering 2015;87:381-388

The next that you have to understand is that the solar collector can not utilize the direct normal solar irradiation, but the part of this which is verticla to the collector aperture. Practically, you can say that the collector aperture is horizontal and so the collector can utilize the product of DNI and cos(θ). So, the net available solar irradiation is DNI*cos(θ). 

Moreover, this part of the solar irradiation reches in the reflector, where there are some losses becuase of the not ideal reflections. The coefficicnt ρ (~0.93 usually) makes a decrease of about 7% for example.

After the reflection, the majority of the rays reaches to the receiver. About 95% of the rays reaches there, but it is something that changes with the insident angle ( θ). We can assume that the 95% of the rays reaches in the idela case (γ=0.95 -intercept factor), 

The next loss i the cover, where a similar loss of about 5-10% is made (τ=92%). The last loss is because the absorber do not capture all teh solar rays, but about the 90% of them (a=90%).

As optical efficicny we can say that: ηopt=ρ*γ*τ*α*Κ(θ)

Where Κ(θ) is teh incident angle modifier. This parameter includes the losses via cos(θ) which are described. Moreover there are end losses (especially when the sun is low in the winter) and cosinus losses because of various errors. Generally the K has this format:

K(θ)=cos(θ)+k0*θ+k1*θ2+...

where k1,k2,... are determined separately for every collector. There are some equations for K but they are theretical, you can found some of them in literature

Sheldon M. Jeter, Dina I. Jarrar, Safwat A. Moustafa, Geometrical effects on the performance of trough collectors, Solar Energy, Volume 30, Issue 2, 1983, Pages 109-113

The next step is to determine the solar heat flux distribution over teh absorber. The better way is to use a software (tracepro or solidworks or any other). There are some literature curves, but these are for secific cases.

Dear Prof. Khalil,

I am so thankful for your consideration in my question. I, for sure, will download the publication and study that. Thank you again. 

Dear Dr. Bellos, 

I don't know how to express my deepest sense of gratitude to your perfect and comprehensive answer. Specifically what I don't understand about the optical efficiency is the parameters affecting intercept factor and the way they are considered and determined. Furthermore, I also have trouble understanding the effects of incidence angle on the optical loss. I would be grateful if you could help me through.

Sincerely

Dear Amin,

Think that there are 100 solar rays that reach to the reflector. If you assume that ρ=0,9, then 90 rays will be reflected to the receiver. However, because there are errors in the tracking system, because the parabolic shape is not perfect in reality, because the receiver is not so great as to absorb all teh reflected rays, some of them are lost and do not reach to the receiver. The number of these lost rays are expreses by the intercept factor. If you suppose again 90% intrect factor, the final number of rays that will reach to the receiver is 0.9*90=81 rays. This parameter is not easily calculated. You can make assumptions if there is no great importance in teh study or to find together all the optical efficiency from literature. I calculate this parameter with solidworks. For example see the next publication:

http://www.sciencedirect.com/science/article/pii/S2213138816300169

For the effects in the incident loss, I will say that there are 2 many effects, the cos(θ) and the end losses. The end losses are small effects in long collectors and you can read the next :

http://www.sciencedirect.com/science/article/pii/S0306261913008969

The losses from cos(θ) are expalined by teh solar position. In teh summer, ehwer the θ close to 0 in the solar noon, more radiation is given to the collector. In the winter, where the solar position is down, the θ is greater and the product DNI*cos(θ) is lower. Try to make some calculations by hand and you will understand. In any case to remember that the collector can utilize the solar beam irradiation in horizontal plane approximatelly. 

Dear Dr. Bellos, 

I must say that you are among those few who have got the concept of PTCs in as much as you can expain it simply to me. And as Einstein says: "if you cant explain it simply enough then you have not realized it well enough". 

Once again I would like to thank you for your perfect answers.

Dear Amin,

Answers from Dr. Evangelos Bellos was very clear! 

This answer was also interesting for me!